Even degree large isogeny computation

I want to calculate $2^{372}$ and $3^{239}$ degree isogenies in Sage, but what I noticed is that using EllipticCurveIsogeny function does not finish executing for such large degree isogenies. There is already a partial answer by Dan Fulea here, but it does not seem to work for the even degree isogenies, and also for the above mentioned large degree odd isogeny as there is no chaining of small degree ones.

where R has degree $2^{372}$. How can one chain 2 and 3 isogenies compute high degree isgonies that are powers of 2 and 3?. Any ideas how to do these efficiently in Sage so that the computations finish a "reasonable" time?

Here is a solution working for both even and odd order of the kernel.
Of course, if the kernel is a "big" composite number,
it is a good idea to split it into factors,
and use the composition isogeny, this telescopical way to proceed is better.

If i understand the situation and the task, things are as follows:

We initialize an elliptic curve $E=E_0$ over a big field $F$,
and declare some rational points, first of all
$R=R_0$ is a point of order $N=3^{239}$ defining a subgroup of rational points, in our case

sage: R.order().factor()
3^239

and then we pick two further points, phiP and phiQ,
on the curve.

Let $R_1=3R$, $R_2=9R$, and so on, $R_n=3^nR$. In our case, $R_N=R_{239}=0$, $N=239$, and let $R'$ be the last
non-zero point in the chain $R_0, R_1,R_2,\dots\ $ So
$$R'=R_{N-1}\ .$$

We consider the isogeny $\phi$ from $E=E_0$ to its
codomain $E'=E_1$, in code EE, which kills $R'$ (and its multiples).

This time the code will not use a class, since it seems to be not so easy to
make transparent the computational idea. Instead, we use plain direct
functional programming.

We will insert in a list the points $R_0, R_1, \dots, R_{N-1}$,
then construct pointwise isogenies w.r.t. the elements in the list,
as long as the list is not empty. All points in the list are mapped
through the isogeny into the new curve.

which give an excellent unifying point of view for both even and odd cases.

"Small" digression. The above article gives only the formula for the $x$--component,
to get the one for the $y$--component, one can proceed as follows.
(The following applies only for the two-torsion case.)
Let us start with the elliptic curve
$E$ given by $y^2=x^3 +ax+b$, which should have a $2$-torsion point $(s,0)$ (over the tacitly chosen definition field).
Then we get rid of $b$, using $s$ instead, $-b=s^3+as$. We also know the equation for the codomain
of the to-be-constructed isogeny $\phi$, all the data put together:
\begin{align}
E\ :\ y^2 &= x^3+ax+b\ ,\\
y^2 &= (x^3-s^3) + a(x-s)\ ,\\
t &= 3s^2 +a\ ,\\
w &= st\ ,\\
E'\ :\ Y^2 &= X^3 + (a-5t)X +(b-7w)\ ,\\
(x,y)&\to (X,Y)=\phi(x,y) \ ,\\
X &= x+\frac t{x-s}\ ,\\
Y &=?\ .
\end{align}
To obtain the value for $Y$, for the second (affine) component in the definition of $\phi$,
we have the Ansatz that $Y$ should be of the shape $Y = yP/(x-s)^2$, where $P$ is a rational function of $x$.
So we only have to detect the $P$. The following sage code does the job:

Note that taking the derivative of X to get Y also works for the
odd-degree-torsion case!
(This was already somehow in my blood before, but i could not find any explicit reference,
maybe i have to write a one page article exactly on this small point, it is too important
structurally, mnemotechnically, and for an implementation as here. Same applies in the odd degree case.)

So let us also test the odd degree torsion case. Same curve, an other $R$.